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My familiarity with concepts related to derived categories is only tangential, and little by little I intend to get more comfortable with them. I was playing around with Caldararu's introduction to the topic, and looking up various things on the web.

Here is my question (that I have every confidence is trivial for experts):

On the wiki page on Verdier duality http://en.wikipedia.org/wiki/Verdier_duality it says the following: Let $F$ be a field, and $X$ a finite dimensional (dimension is defined here cohomologically, but for our purposes a finite dimensional manifold will do) locally compact space.

In the part about Poincare duality, it says: $H^k(X,F)=[F,X[k]]$. What is the interpretation of this notation? As I see it, $[F,X[k]]$ means $Hom(F,X[k])$ in the derived category. But this means that $X$ is seen as a complex. How? And why would $Hom(F,X[k])$ equal $H^k(X,F)$?

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    $\begingroup$ I guess that there is a typo and that $[F,X[k]]$ should be understood as $[F,F[k]]$. $\endgroup$
    – DamienC
    Commented Jun 23, 2011 at 20:27
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    $\begingroup$ Why is $[F,F[k]]=H^k(X,F)$? $\endgroup$
    – James D. Taylor
    Commented Jun 23, 2011 at 20:30
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    $\begingroup$ Very shortly, for any sheaf $S$ (i.e. a complex of sheaves concentrated in degree $0$), $[F[-k],S]=H^k(X,S)$. $\endgroup$
    – DamienC
    Commented Jun 23, 2011 at 20:37
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    $\begingroup$ Thanks! You mean $[S[-k],S]$, right? Because $H^k(X,S)$ doesn't depend on $F$. $\endgroup$
    – James D. Taylor
    Commented Jun 23, 2011 at 20:40
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    $\begingroup$ I do really mean $F$, because I ma working with the derived category of $F$-modules (as in the wiki). But you can just take $F=\mathbb{Z}$ if you like. $\endgroup$
    – DamienC
    Commented Jun 23, 2011 at 20:54

2 Answers 2

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I think here is how one should understand the last paragraph of the wiki.

Consider $f:X\to pt$. We have (all functors are derived and my $Hom$ are sheaf $Hom$) $$ Hom_{pt}(f_!F,F)=f_*Hom_X(F,f^!F) $$

$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the dual of the cohomology with compact support of $X$.

$Hom_X(F,f^!F)= \Gamma(f^!F)$ and thus the r.h.s. is $f_*(\Gamma(f^!F))=\Gamma(X,f^!F)$, the homology of $X$.

EDIT : to answer precisely the question I include a summary of the comments.

  1. there is a typo in the wiki: $[F,X[k]]$ should be understood as $[F,F[k]]$.

  2. for any sheaf of $F$-modules $S$ (concentrated in degree 0), $[F[−k],S]=H^k(X,S)$.

  3. Contrary to what is claimed in the wiki, there is no duality between $H^k(X,F)=[F[−k],F]$ and $H_k(X,F)=[F[−k],D_X]$ (where $D_X:=f^!F$ for $f:X\to pt$). The duality is, either between $H^k_c(X,F)$ and $H_k(X,F)$, or between $H^k(X,F)$ and $H_k^{BM}(X,F)$. And as far as I understand $H^*(X,S)$ is dual to $H_c^*(X,S^\vee)$.

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  • $\begingroup$ I'm still a little confused. Under some conditions, in Poincare duality we have $D_X=F[-n]$, right? So $H_k(X,F)=Hom(F[-k],F[-n])=Hom(F,F[k-n])=H^{k-n}(X,F)=0$. I'm doing something wrong. $\endgroup$
    – James D. Taylor
    Commented Jun 23, 2011 at 21:11
  • $\begingroup$ Oh, I guess you're saying that $Hom(F,F[a])\neq H^a(X,F)$, but rather $H^a(X,F)=Hom(O_X,F[a])$. But still, how does one get from $Hom(F,F[k-n])$ to the desired $H^{n-k}(X,F)$ dual? I guess $Hom(F,F[k-n])=Hom(O_X,F[k-n]$ dual $)$. Now what? $\endgroup$
    – James D. Taylor
    Commented Jun 23, 2011 at 21:17
  • $\begingroup$ Aha! Here's what's missing. For some reason: $Hom(O_X,C)\cong$ to the dual of $H^(X,$ dual of $C)$. If I can only figure out why that's true, I'll be satisfied. $\endgroup$
    – James D. Taylor
    Commented Jun 23, 2011 at 21:19
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    $\begingroup$ Sorry. The point is that there is another mistake in the wiki: Poincaré duality is - either between $H^*$ and $H^{BM}_*$ (Borel-Moore homology) - either between $H_c^*$ and $H_*$ - not between $H^*$ and $H_*$. Then one has $H^k(X,F)=[F[-k],F]$, $H_k(X,F)=[F[-k],D_X]$, and $H^k_c(X,F)$ is given by the formula in you comment. $\endgroup$
    – DamienC
    Commented Jun 23, 2011 at 22:22
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    $\begingroup$ @euklid345: nobody's complaining (IMO spotting a typo and/or a mistake is not complaining). $\endgroup$
    – DamienC
    Commented Jan 11, 2012 at 8:02
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I just revamped what was written. Perhaps now it is more understandable:New Wiki Entry on Verdier duality.

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